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In the xy coordinate plane, line L and line K intersect at [#permalink]
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06 May 2010, 11:31
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In the xy coordinate plane, line L and line K intersect at the point (4,3). Is the product of their slopes negative? (1) The product of the xintersects of lines L and K is positive. (2) The product of the yintersects of lines L and K is negative.
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Last edited by Bunuel on 24 May 2012, 11:49, edited 1 time in total.
Edited the question and added the OA



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In the xy coordinate plane, line L and line K intersect at [#permalink]
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06 May 2010, 12:24
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In the xy coordinate plane, line L and line K intersect at the point (4,3). Is the product of their slopes negative?We have two lines: \(y_l=m_1x+b_1\) and \(y_k=m_2x+b_2\). The question: is \(m_1*m_2<0\)? Lines intersect at the point (4,3) > \(3=4m_1+b_1\) and \(3=4m_2+b_2\) (1) The product of the xintersects of lines L and K is positive. Now, one of the lines can intersect xaxis at 0<x<4 (positive slope) and another also at 0<x<4 (positive slope), so product of slopes also will be positive BUT it's also possible one line to intersect xaxis at 0<x<4 (positive slope) and another at x>4 (negative slope) and in this case product of slopes will be negative. Two different answers, hence not sufficient. But from this statement we can deduce the following: xintersect is value of \(x\) for \(y=0\) and equals to \(x=\frac{b}{m}\) > so \((\frac{b_1}{m_1})*(\frac{b_2}{m_2})>0\) > \(\frac{b_1b_2}{m_1m_2}>0\). (2) The product of the yintersects of lines L and K is negative. Now, one of the lines can intersect yaxis at 0<y<3 (positive slope) and another at y<0 (positive slope), so product of slopes will also be positive BUT it's also possible one line to intersect yaxis at y<0 (positive slope) and another at y>3 (negative slope) and in this case product of slopes will be negative. Two different answers, hence not sufficient. But from this statement we can deduce the following: yintercept is value of \(y\) for \(x=0\) and equals to \(y=b\) > \(b_1*b_2<0\). (1)+(2) \(\frac{b_1b_2}{m_1m_2}>0\) and \(b_1*b_2<0\). As numerator in \(\frac{b_1b_2}{m_1m_2}>0\) is negative, then denominator \(m_1m_2\) must also be negative. So \(m_1m_2<0\). Sufficient. Answer: C. In fact we arrived to the answer C, without using the info about the intersection point of the lines. So this info is not needed to get C. For more on coordinate geometry check the link in my signature. P.S. This question can be easily solved by drawing the lines without any calculations.
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Re: Interesting Line Eqn DS problem [#permalink]
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19 Jul 2010, 07:29
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line L: y1 = m1x1 + b1; xintercept = b1/m1; yintercept = b1 line K: y2 = m2x2 + b2; xintercept = b2/m2; yintercept = b2 (4,3) lies on both. 3 = 4m1 + b1 = 4m2 + b2 Is m1m2 < 0? 1. (b1/m1)*(b2/m2) > 0 b1b2/m1m2 > 0 If b1b2 > 0, m1m2 > 0 If b1b2 < 0, m1m2 < 0 NOT SUFFICIENT. 2. b1b2 < 0 NOT SUFFICIENT. Together, b1b2 < 0 and m1m2 < 0. SUFFICIENT. Answer is C.
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Re: GMAT PREP (DS) [#permalink]
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25 Sep 2010, 11:27
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The equation of a line can be written as : \(\frac{x}{a} + \frac{y}{b} = 1\) Here a is the Xintercept and b is the Yintercept The slope of this line is b/a Let the lines be : \(\frac{x}{a_1} + \frac{y}{b_1} = 1\) \(\frac{x}{a_2} + \frac{y}{b_2} = 1\) We know both lines pass through (4,3) So we know that \(\frac{4}{a_1} + \frac{3}{b_1} = \frac{4}{a_2} + \frac{3}{b_2} = 1\) The information about (4,3) doesn't tell us much about the signs of a1,a2,b1,b2 Question is if Slope1*Slope2 =(b1 b2) / (a1 a2) < 0 (1) a1 * a2 > 0 Depends on the sign of b1*b2 (2) b1 * b2 < 0 Depends on the sign of a1*a2 (1) + (2) : Clearly sufficient to say yes. The slope product is always < 0.
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Re: GMAT PREP (DS) [#permalink]
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27 Sep 2010, 07:39
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under timed schedule, i considered using a quick graphical sketch and that helped. Let M1 and M2 represent the gradients of lines K and L (1) from sketch (i), M1*M2 < 0 from sketch (ii), M1*M2 > 0 INSUFFICIENT (2) from sketch (i), M1*M2 < 0 from sketch (iii), M1*M2 > 0 INSUFFICIENT Combining (1) and (2): ONLY sketch i) satisfies > SUFFICIENCY option C is therefore correct.
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Re: GMAT PREP (DS) [#permalink]
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In the xycoordinate plant, line t and k intersect at (4,3). Is the product of their slopes negative? 1. Product of the x intercepts of the line t & k is positive 2. Product of the y intercepts of the line t & k is negative Sol: To draw a line with +ve slope through a point(x1, y1) on a 2D plane; draw two lines passing through the point perpendicular to each other with one line parallel to xaxis and the other parallel to yaxis(depicted by red lines in the images). These redlines make 4quadrants with origin(x1,y1). Any line that passes through this origin and lies on 1st and 3rd quadrant will have +ve slope. Any line passing through the point and lying on the 2nd and 4th quadrants will have negative slope. Attached images show two possible scenarios for St1 and St2 making the statements insufficient individually. Last image shows the only possible scenario for the lines, making both the statements sufficient together. St1: Insufficient. Attachment:
product_slopes_of_two_lines_3_n_4_St1.PNG [ 14.96 KiB  Viewed 10336 times ]
St2: Insufficient. Attachment:
product_slopes_of_two_lines_3_n_4_St2.PNG [ 15.7 KiB  Viewed 10330 times ]
Combined: Sufficient. Attachment:
product_slopes_of_two_lines_3_n_4_Combined.PNG [ 12.94 KiB  Viewed 10305 times ]
Ans: "C"
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Re: GMAT PREP (DS) [#permalink]
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27 Apr 2011, 19:13
(1) So they both have +ve xintercept or ve x=intercept This is not sufficient, as can be seen from diagram, both can have +ve slope or 1 can have ve and 1 +ve slope (2) So they both intersect yaxis on different sides of XAxis This is also not sufficient, as can be seen from diagram, both can have +ve slope or 1 can have ve and 1 +ve slope (1) and (2) They are sufficient, as only the option with two lines on right of yaxis is possible. Hence products of slopes is negative. Answer  C
Attachments
Lines.JPG [ 16.69 KiB  Viewed 9516 times ]
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Re: GMAT PREP (DS) [#permalink]
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26 May 2012, 00:11
Bunuel wrote: LM wrote: Please explain... In the xy coordinate plane, line L and line K intersect at the point (4,3). Is the product of their slopes negative?We have two lines: \(y_l=m_1x+b_1\) and \(y_k=m_2x+b_2\). The question: is \(m_1*m_2<0\)? Lines intersect at the point (4,3) > \(3=4m_1+b_1\) and \(3=4m_2+b_2\) (1) The product of the xintersects of lines L and K is positive. Now, one of the lines can intersect xaxis at 0<x<4 (positive slope) and another also at 0<x<4 (positive slope), so product of slopes also will be positive BUT it's also possible one line to intersect xaxis at 0<x<4 (positive slope) and another at x>4 (negative slope) and in this case product of slopes will be negative. Two different answers, hence not sufficient. But from this statement we can deduce the following: xintersect is value of \(x\) for \(y=0\) and equals to \(x=\frac{b}{m}\) > so \((\frac{b_1}{m_1})*(\frac{b_2}{m_2})>0\) > \(\frac{b_1b_2}{m_1m_2}>0\). (2) The product of the yintersects of lines L and K is negative. Now, one of the lines can intersect yaxis at 0<y<3 (positive slope) and another at y<0 (positive slope), so product of slopes will also be positive BUT it's also possible one line to intersect yaxis at y<0 (positive slope) and another at y>3 (negative slope) and in this case product of slopes will be negative. Two different answers, hence not sufficient. But from this statement we can deduce the following: yintercept is value of \(y\) for \(x=0\) and equals to \(x=b\) > \(b_1*b_2<0\). (1)+(2) \(\frac{b_1b_2}{m_1m_2}>0\) and \(b_1*b_2<0\). As numerator in \(\frac{b_1b_2}{m_1m_2}>0\) is negative, then denominator \(m_1m_2\) must also be negative. So \(m_1m_2<0\). Sufficient. Answer: C. In fact we arrived to the answer C, without using the info about the intersection point of the lines. So this info is not needed to get C. For more on coordinate geometry check the link in my signature. P.S. This question can be easily solved by drawing the lines without any calculations. great explanation.... i just plotted the lines on a coordinate figure.... st 1: product of xintercepts are positive, so from point (4,3) in first quadrant we could have two possibilities: a) 1st line  negative slope and a positive xintercept and 2nd line  positive slope and positive xintercept (yintercepts positive  line 1 and negative = line 2) b) 1st line  positive slope and a positive xintercept and 2nd line  positive slope and positive xintercept (yintercepts negative) hence 2 cases are possible... not sufficient st 2: approach the statement with the same philosophy as in st 1 and arrived at two cases, a) 1st line  negative slope and a positive yintercept and 2nd line  positive slope and negative yintercept (xintercepts positive) b) 1st line  positive slope and a negative yintercept and 2nd line  positive slope and positive yintercept (xintercepts positive  line 1 and negative  line 2) hence 2 cases are possible... not sufficient from both the statements, we can arrive at a unique case case a) from st 1 and case a) from st 2 answer therefore is C the cases look very much conceivable on a graph plot.



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In the xy coordinate plane, line L and line K intersect at the p [#permalink]
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10 May 2013, 11:58
M8 wrote: In the xycoordinate plane, line l and line k intersect at the point (4,3). Is the product of their slope negative? 1) The product of the xintercepts of lines l and k is positive. 2) The product of the yintercepts of lines l and k is negative. I cracked this question right. But these types of questions are real horror for me. Guys please elaborate your solution for me. Thank you. I'll post the OA later. Let X1 and X2 are the X intercepts for the lines l and K. Calculate slope for L [(4,3) and (X1, 0)] and m [ (4,3) and (X2, 0)] such that X1*X2=+VE Similarly take Y1 and Y2. and calculate slope for L [(4,3) and (0, Y1)] and m [ (4,3) and (0, Y2)] such that Y1*Y2=VE Now combine these two statement and calculate slope for L [(0, Y1) and (X1, 0)] and m [ (0, Y2) and (X2, 0)] Slope of L*M = Y1/(X1)*Y2/(X2) = Y1*Y2/ X1*X2. Given that Y1*Y2= VE and X1*X2=+VE. Hence Slope of L*M = VE Answer C.



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Re: In the xy coordinate plane, line L and line K intersect at [#permalink]
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27 Jun 2013, 21:28



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Re: In the xy coordinate plane, line L and line K intersect at [#permalink]
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25 Nov 2014, 07:01
shrouded1 wrote: The equation of a line can be written as :
\(\frac{x}{a} + \frac{y}{b} = 1\)
Here a is the Xintercept and b is the Yintercept
The slope of this line is b/a
Let the lines be :
\(\frac{x}{a_1} + \frac{y}{b_1} = 1\) \(\frac{x}{a_2} + \frac{y}{b_2} = 1\)
We know both lines pass through (4,3) So we know that \(\frac{4}{a_1} + \frac{3}{b_1} = \frac{4}{a_2} + \frac{3}{b_2} = 1\) The information about (4,3) doesn't tell us much about the signs of a1,a2,b1,b2
Question is if Slope1*Slope2 =(b1 b2) / (a1 a2) < 0
(1) a1 * a2 > 0 Depends on the sign of b1*b2
(2) b1 * b2 < 0 Depends on the sign of a1*a2
(1) + (2) : Clearly sufficient to say yes. The slope product is always < 0. Did this way using the intercept form of a line. Have made an observation that using intercept equations in gmat questions significantly reduces the time to solve a question. Thanks. A thumb rule I use in coordinate geometry for lines: 1. Use intercept form x/a + y/b = 1  If intercepts are given, for area questions usually, for ladder against wall type questions, triangle questions2. Use slope form y=mx+c  when slope is implicit or explicitly stated in the question 3. Use one point slope form yy1=m(xx1)  when we know slope and line passes through a point4. Use 2 point form yy1/y2y1 = xx1/x2x1  when line passes through two points5. When all else fails we resort to pure algebra of a linear equation ax+by+c=0
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Re: In the xy coordinate plane, line L and line K intersect at [#permalink]
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13 Nov 2016, 16:29
Can the question be solved the following way? Or, am I just lucky with the solution particular to this question?
Slope of a line = \(\frac{rise}{run}\) = \(\frac{yintercept}{xintercept}\)
With this in mind,
(1) The product of the xintercepts of lines \(l\) and \(k\) is positive
Translation:
\(\frac{y}{x}\) = \(+\) or \(\frac{y}{x}\) = \(\)
Insufficient
(2) The product of the yintercepts of lines \(l\) and \(k\) is negative
Translation:
\(\frac{y}{x}\) = \(+\) or \(\frac{y}{x}\) = \(\)
Insufficient
(1) & (2):
\(\frac{y}{x}\) = \(\) is the only answer in common and therefore, sufficient.
Please advise,
Thank you!



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Re: In the xy coordinate plane, line L and line K intersect at [#permalink]
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